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Let $X,Y$ be smooth irreducible algebraic varieties over $\mathbb{C}$, let $\pi: X \to Y$ be a morphism which is injective and surjective on closed points of $X,Y$. Then how to show $\pi$ is an isomorphism of $X,Y$ as varieties?

This result is not true if $X,Y$ are not smooth. One can consider the example $\mathbb{P}^1 \to C$, where $C$ is a plane curve defined by $y^2-x^3=0$, because the map $t \mapsto (t^2,t^3)$ satisfies the condition, but $\mathbb{P}^1$ cannot isomporhic to $C$ which is singular at origin.

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    $\begingroup$ Use (a special case of) Zariski's Main Theorem. $\endgroup$
    – anon
    Commented Mar 16, 2013 at 19:46
  • $\begingroup$ The map must be birational. Since Y is normal, it has to be an isomorphism. $\endgroup$
    – Ray Hoobler
    Commented Mar 16, 2013 at 21:15
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    $\begingroup$ It is in EGA IV, 3ème partie, Théorème 8.12.6, page 45, it is called " ,,Main Theorem'' de Zariski." It says that your map decomposes into an open immersion, followed by a finite morphism. In your case, you should be able to see that the both are isomorphisms. $\endgroup$
    – Jérémy Blanc
    Commented Mar 16, 2013 at 21:22
  • $\begingroup$ @Ray Hoobler, could you give more details on how the map is birational? $\endgroup$
    – Li Yutong
    Commented Mar 16, 2013 at 21:54
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    $\begingroup$ Since it is injective, it is birational with its image (compute the degree of field extension: it is $1$). Because it is surjective, it is birational. $\endgroup$
    – Jérémy Blanc
    Commented Mar 17, 2013 at 9:58

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One way to prove this is to show that for each closed point $x\in X$ the differential $T_\pi(x):T_xX\rightarrow T_{\pi(x)}Y$ is surjective, that is $\pi$ is a smooth morphism.

Let $\pi:X\rightarrow Y$ be a morphism or relative dimension $r$ of smooth varieties over an algebraically closed field. Assume that the relative cotangent sheaf $\Omega_{X/Y}$ is locally free of rank $r$ on $X$. We have an exact sequence $$\pi^{*}\Omega_{Y}\rightarrow\Omega_{X}\rightarrow\Omega_{X/Y}\mapsto 0.$$ Let $k(x)$ be the residue field at a closed point $x$. Tensorizing we get
$$\pi^{*}\Omega_{Y}\otimes k(x)\rightarrow\Omega_{X}\otimes k(x)\rightarrow\Omega_{X/Y}\otimes k(x)\mapsto 0.$$ Since $X$ and $Y$ are smooth and $\Omega_{X/Y}$ is locally free of rank $r$ these three vector spaces are of dimension $dim(Y),dim(X),r$ respectively. So the first map is injective and we have $$0\mapsto\pi^{*}\Omega_{Y}\otimes k(x)\rightarrow\Omega_{X}\otimes k(x)\rightarrow\Omega_{X/Y}\otimes k(x)\mapsto 0.$$ For any closed point $x\in X$ we have $k(x)\cong k$. Therefore we can identify the injective map $\pi^{*}\Omega_{Y}\otimes k\rightarrow\Omega_{X}\otimes k$ with the map between the cotangent spaces $\mathfrak{m}_y/\mathfrak{m}^2_y\rightarrow\mathfrak{m}_x/\mathfrak{m}^2_x$ where $y = \pi(x)$. Dualizing we have that the differential $T_\pi(x):T_xX\rightarrow T_yY$ is surjective. Therefore $\pi$ is a smooth morphism.

In particular, if any fiber of $\pi$ is just one point, then $\Omega_{X/Y}$ is locally free of rank $r = 0$ and $\pi$ is a smooth morphism of relative dimension zero. Finally, since $\pi$ is surjective it is an isomorphism.

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  • $\begingroup$ Thank you! Why $\Omega_{X/Y}$ is locally free when the fibre of $\pi$ is just one point? Your argument let me thing that flatness of $\pi$ is enough to show the isomorphism --- is that right? $\endgroup$
    – Li Yutong
    Commented Apr 2, 2014 at 14:31
  • $\begingroup$ I am assuming that the fiber of $\pi$ is reduced. Therefore $\Omega_{X/Y}$ is locally free. In general flat is not enough. For instance project a conic $C$ in $\mathbb{P}^2$ from a point $p\notin C$. You have two points $x,y\in C$ where the lines through $p$ are tangent. The projection $\pi:C\rightarrow\mathbb{P}^1$ is flat. However $T_{\pi}(x):T_xC\rightarrow T_{\pi(x)}\mathbb{P}^1$ is not surjective, and $T_{\pi}(y)$ is not surjective as well. As points of the fiber of $\pi$ both $x,y$ have multiplicity two. $\endgroup$
    – Puzzled
    Commented Apr 2, 2014 at 16:17
  • $\begingroup$ Thank you again! Is the statement that "The fibre of $\pi$ is reduced then $\Omega_{X/Y}$ is locally free" somewhere in standard text book? I have never seen it before:( $\endgroup$
    – Li Yutong
    Commented Apr 3, 2014 at 0:33
  • $\begingroup$ Besides, I might be understand wrong, but in your example, isn't the fibre of $\pi$ always two points? So it does not satisfy the requirement that $\pi$ is injective? $\endgroup$
    – Li Yutong
    Commented Apr 3, 2014 at 0:34
  • $\begingroup$ I thought flatness is okay because if $\pi$ is proper, then injective of $\pi$ would imply $\pi$ to be finite, and flatness would imply it is an isomorphism (because locally, it is a rank one free module). Is there anything wrong in that argument? $\endgroup$
    – Li Yutong
    Commented Apr 3, 2014 at 0:36

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